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Mendelian Genetics

Jan 96,39

Use the following information to answer the next four questions

The human nose can distinguish up to 10,000 different odours. Airborne odour molecules are

trapped by ciliated olfactory receptors. These receptors initiate impulses to the olfactory bulbs,

which relay the signals to the temporal lobe of the cerebrum for interpretation.

Recently, researchers Linda Buck and Richard Axel of Columbia University have been able to

identify some odour-receptor proteins on the nasal cells of rats. These receptor proteins, embedded

in the cell membranes of ciliated neurons, are thought to bind with specific odour molecules,

initiating a nerve impulse. The researchers have also found the genes responsible for the

production of more than 100 types of odour-receptor proteins.

Jan 96,41

1. The ability to smell some odours and not others is an inherited trait. This trait is an example of

A. incomplete dominance

B. crossing over

C. a phenotype

D. a genotype

Jan 96,42

2. If a gene responsible for the production of an odour-receptor protein underwent a mutation, one result

might be

A. a decrease in the ability to smell a specific odour

B. an increase in the ability to smell a variety of other odours

C. a decrease in protein synthesis within all olfactory neurons

D. an increase in threshold levels of stimulation for receptor neurons

Jan 96,43

Use the following information to answer the next two questions

Some researchers are investigating the hypothesis that feeding cow’s milk to children less than

nine months old could play a role in the development of type I (juvenile) diabetes mellitus in those

genetically prone to the disease. A child who has inherited certain genes synthesizes antibodies in

response to a particular fragment of a protein found in cow’s milk. These antibodies attack the

milk-protein fragment and some of the child’s cells. After a child reaches the age of nine months,

the immune system develops sufficiently to distinguish between the milk-protein fragment and the

child’s cells.

3. If a child who is genetically prone to juvenile diabetes has consumed cow’s milk, it is likely that the

child’s blood contains antibodies that attack the

A. sensory cells of the hypothalamus

B. glycogen storage cells of the liver

C. secretory cells of the thyroid

D. islet cells of the pancreas

Jan 96,44

4. A correct inference is that the antibodies produced in response to cow’s milk are

A. nucleic acids containing a sequence of amino acids

B. nucleic acids containing a sequence of nucleotides

C. proteins containing a sequence of amino acids

D. proteins containing a sequence of nucleotides

Jan 96,45

Use the following information to answer the next three questions

The Romanovs, the Russian royal family, were shot dead during the Russian Revolution in 1918.

The family included Czar Nicholas II and his wife Alexandra; their daughters, Olga, Maria,

Tatiana, and Anastasia; and their son Alexis, who was a hemophiliac.

In 1992, scientists in Russia started work on the identification of skeletons thought to be the

remains of the Czar, his wife, and three of their daughters. The researchers crushed and partially

dissolved some bones to extract DNA.

To determine which skeletons belonged to the Romanovs, the researchers analyzed samples of

mitochondrial DNA. Mitochondrial DNA is passed on only from mother to child. Some parts of

mitochondrial DNA are very stable and mutate only once every 6,000 years. The researchers also

obtained a sample of mitochondrial DNA from Prince Philip of England. His maternal

grandmother was Alexandra’s sister.

5. Which observation correctly accounts for the fact that mitochondrial DNA is passed on only from

mother to child?

A. eggs contain mitochondria and sperm do not

B. mitochondrial DNA is transcribed from a template on the XY chromosome set

C. mitochondrial DNA is transcribed from a template on the XX chromosome set

D. an egg contributes both its nucleus and its cytoplasm to a zygote, while a sperm

contributes only its nucleus

Jan 96,46

6. From which people would samples of mitochondrial DNA be nearly identical

A. Prince Philip, Czar Nicholas II, and Alexandra

B. Alexis, Prince Philip and Czar Nicholas II

C. Czar Nicholas II, Alexandra, and Alexis

D. Alexandra, Alexis, and Prince Philip

Jan 96,8

Numerical Response

1. Czar Nicholas II was not a hemophiliac. His wife, Alexandra, was a carrier for hemophilia. What was

the probability that any of their daughters was a carrier for hemophilia?

(Record your answer as a value from 0 to 1, rounded to two significant digits, in the numerical-

response section of the answer sheet.)

Answer: ___________

Jan 96,47


William Jia, a researcher at the University of British Collumbia, thinks a genetically altered

Herpes virus may one day be used to destroy gliomas, a type of brain cancer that attacks the

frontal lobes of the cerebrum.

Jia managed to genetically engineer a Herpes virus that replicates only in rapidly dividing cancer

cells and thus destroys only those cells. Jia handicaps the Herpes virus by deleting the genes that

code for an enzyme needed for viral replication. This prevents the virus from replicating in

normal, non-dividing brain cells, but enable it to continue replicating in cancerous brain cells. As

the virus replicates, the rapidly dividing cancer cells are killed.

7. The deletion of genes from the Herpes virus likely involves the use of

A. biopsies and chorionic villus sampling

B. karyotyping and pedigree analysis

C. amniocentesis and chromosome mapping

D. restriction enzymes and ligases

Jan 96,48

8. After the destruction of the cells in a brain tumour, the altered Herpes virus will likely

A. change into a form capable of producing the enzymes necessary for the reproductive

cycle of the virus

B. spread to other areas of the body and attack other types of cancer cells that may be

present

C. enter a dormant stage because enzymes required for replication are absent

D. spread to a nearby region of the brain and begin normal replication

Jun 96,27

Use the following information to answer the next question.

When conducting his research into the genetics of pea plants, Gregor Mendel crossed true-

breeding, tall pea plants with true-breeding, short pea plants.

The offspring of this P1 generation, the F1 generation, were interbred and produced the F2

generation. The height of each plant in the F1 and the F2 generations was recorded.

9. In pea plants, if tall is dominant over short, which graph best represents the expected distribution of

offspring in the F2 generation?

Jun 96,2

Use the following information to answer the next question

Mendel also found that in peas, yellow seeds (Y) are dominant over green seeds (y), and round

seed shape (R) is dominant over wrinkled seed shape (r). the alleles for these traits are located on

independent chromosomes.

Numerical Response

2. When a YyRr parent and a YYrr parent are crossed, what percentage of their offspring are expected to

have yellow seed-coat colour and wrinkled seed shape?

(Record your answer in the numerical-response section of the answer sheet.)

Answer: _____________ %

Jun 96,28


In his work with the fruit fly (Drosophila), Thomas Morgan bred normal red-eyed female flies

with white-eyed males. He bred enough flies to have acceptable statistical significance and

reliability. The F1 generation were all red-eyed. All the females of the F2 generation had red eyes,

but one-half of the males of the F2 generation had white eyes.

10. A conclusion that may be drawn from Morgan’s work is that the gene for white eyes is located on the

A. Y chromosome and is recessive

B. Y chromosome and is dominant

C. X chromosome and is recessive

D. X chromosome and is dominant

Jun 96,29


In Drosophila, the genes for eye colour (pr), wing shape (vg), and body colour (eb) are all found

on the same chromosome. The following crossover frequencies for these genes were determined

by experimentation.

Genes Crossover Frequency

pr and vg 12.5%

pr and eb 6.0%

vg and eb 18.5%

11. Which is the correct sequence of the genes pr, vg, and eb on the chromosome?

A. pr–vg–eb

B. vg–eb–pr

C. eb–pr–vg

D. eb–vg–pr

Jun 96,31


12. The pedigree indicates that the trait for opalescent dentin is controlled by a gene that is

A. X - linked

B. recessive

C. dominant

D. codominant

Jun 96,3

Numerical Response

3. If individual III 1 and individual III 2 had another child, what is the probability that this child would

be a female with opalescent dentin?

(Record your answer as a value from 0 to 1, rounded to two decimal places, in the numerical-


Answer: __________

A Pedigree Showing the Incidence of Opalescent Dentin in a Family

This pedigree shows the incidence of opalescent dentin in a family. People with this trait have insufficient

enamel on their teeth. The dentin inside the teeth shows through, giving the teeth a mottled, pearl-like

appearance.

Jun 96,32


13. Which row provides correct information about the genotypes for individuals I 1, II 1, and II 4?

Row Individual I 1 Individual II 1 Individual II 4

A IAi I

AIA or I

Ai ii

B IAIA I

AIA or I

Ai I

Ai

C IAi I

AIA or I

Ai or I

AIB or I

Bi I

Ai

D IAIA I

AIA or I

Ai or I

AIB or I

Bi ii

Jun 96,33

14. Which row correctly shows the probability of individuals II 2 and III 3 having type O blood?

Row

Probability of individual

II 2 having type O blood

Probability of individual

III 3 having type O blood

A 0.00 0.00

B 0.00 0.25

C 0.25 0.00

D 0.25 0.25

Jun 96,34

15. Protanopia, or red colour blindness, is a recessive X-linked characteristic. A woman with normal

colour vision who carries the allele for red colour blindness and a man who is a red colour-blind are

expecting a child. Which statement about the child is likely correct?

A. If the child is a boy, there is a 100% chance that he will be colour blind

B. If the child is a girl, thre is a 100% chance that she will be colour-blind

C. There is a 25% chance that the child will be colour-blind

D. There is a 50% chance that the child will be colour-blind

Jun 96,35


The ability to taste the chemical phenylthiocarbamide (PTC) is controlled by the dominant allele T.

Individuals who have the allele T find that PTC tastes bitter. Homozygous recessive individuals find

that PTC has no taste. In a sample of 320 people, 218 could taste PTC and 102 could not taste PTC.

16. Which row provides the probable percentage of this population with each genotype indicated?

Genotype

Row TT Tt tt

A 15 % 29 % 56 %

B 44 % 37 % 19 %

C 19 % 49 % 32 %

D 23 % 45 % 32 %

Jun 96,36


A plant breeder was trying to produce a true-breeding variety of radishes with an oval shape.

However, each time plants with oval radishes were self-fertilized, only half of the offspring were

oval. The rest of the plants produced either round or long radishes.

17. What would be the expected phenotype ratio if plants with round radishes were crossed with plants

with oval radishes?

A. All round

B. All oval

C. 1 round : 1 oval

D. 1 round : 1 long

Jun 96,4


Numerical Response

4. What was the phenotypic ratio of the offspring ( including all eight possible genotypes) produced by

the parents?

(Record your four-digit answer in the numerical-response section of the answer sheet.)

Answer: __________ __________ __________ __________

Black-straight black-curly white-straight white-curly

Jun 96,5


In some cattle, coat colour and pattern is controlled by an autosomal gene locus that may contain

one of four alleles. The dominance hierarchy for these alleles is

S (Dutch belt) sh (Hereford) s

c (Solid) s (Holstein)

Numerical Response

5. A Dutch belt bull carrying the allele for Hereford was crossed with a pure-breeding Holstein cow.

What is the probability that the offspring would be a Hereford?



Answer: ___________

Jun 96,6

Numerical Response

6. Snapdragons may be tall or short, and they may be red, white, or pink. The allele for short is recessive

and the alleles for colour display incomplete dominance. A plant, heterozygous for both traits, was

self-pollinated and produced 200 offspring. How many of these 200 offspring would be expected to be

both tall and pink/

(Record your answer in the numerical-response section of the answer sheet.)

Answer: ______________ plants

In guinea pigs, black coat (B) is dominant over white coat (b) an straight hair (S) is dominant over curly hair (s). Each

pair of alleles assorts independently.

Two guinea pigs were crossed. The gametes produced by the parents are shown in the Punnett Square.

♂

♀

BS

Bs

bS

bs

Bs

bs

Note: ♂ male

♀ female

Jun 97,36


18. The inheritance pattern of this disease is most likely

A. X-linked recessive

B. X-linked dominant

C. autosomal recessive

D. autosomal dominant

Jun 97,37

19. Which structure or organ would not develop properly in a person with hypohidrotic ectodermal

dysplasia?

A. Muscle

B. Heart

C. Liver

D. Skin

Jun 97,38

20. Which individuals from the pedigree are definitely carriers of the disease?

A. I-1 and II-5

B. I-2 and II-1

C. III-1 and III-4

D. III-2 and III-4

Hypohidrotic ectodermal dysplasia is a rare genetic disease that cause problems with most embryonic

ectoderm development, but does not usually affect the nervous system. Individuals are born apparently

healthy but lack sweat glands.

Note: Carriers have not been identified and all marriage partners are known to be unrelated.

From Discover

Jun 97,39


21. The row that correctly shows the expected ratio of offspring phenotypes of a cross between two dogs

heterozygous for height and tail curl is

Row

Normal Height

Curled Tail

Normal Height

Uncurled Tail

Dwarf

Curled Tail

Dwarf

Uncurled Tail

A. 0 1 1 0

B. 1 1 1 1

C. 1 3 3 9

D. 9 3 3 1

Jun 97,40

22. If a dog that is heterozygous for height ad tail curl is crossed with a dog that is recessive for both traits,

the four possible genotypes for the offspring are

A. HhCc, Hhcc, hhCc, hhcc

B. HhCc, HhCC, hhCc, hhcc

C. HhCc, Hhcc, hhCC, hhcc

D. HhCc, HHCc, hhCc, hhcc

Jun 97,41


Feather colour for Andalusian fowl is governed by incomplete dominance of a pair of alleles.

Fowl may have black, whit, or blue feathers. Blue-feathered birds are heterozygotes. In a

randomly mating population of 400 fowl, there were 49 white-feathered birds.

23. The allele frequencies p (black) and q (white), respectively, are

A. p = 0.3 and q =0.7

B. p = 0.7 and q = 0.3

C. p = 0.35 and q = 0.65

D. p = 0.65 and q = 0.35

Jun 97,4

Numerical Response

7. What is the frequency of the heterozygous genotype in this population of Andalusian fowl?



Answer: ___________

Jun 97,42

24. When a black-feathered hen is mated with a white-feathered rooster, what feather colour will the

offspring have?

A. All will have blue feathers

B. All will have black feathers

C. Some will have black and some will have white feathers

D. Some will black, some will have white, and some will have blue feathers

Jun 97,5

Numerical Response

8. In a cross between a blue-feathered rooster and a white-feathered hen, what percentage of the offspring

are expected to be white-feathered?

(Record your answer as a whole number percentage in the numerical-response section of the answer

sheet.)

Answer: _____________%

Jun 97,43


25. How many people in Generation III are homozygous for blood type?

A. 1

B. 2

C. 3

D. 4

Jun 97,6

Numerical Response


9. Predict the order in which the genes occur on chromosome 2. Use the numbers following the

characteristics to code your answer


(There are two possible ways of recording the answer; either will be acceptable.)

Answer: ____________

The characteristics plexus wings, scabrous eyes, speckled body, and brown eyes are influenced

by genes found on chromosome 2 of the common fruit fly, Drosophila melanogaster.

Crossover frequencies between the genes are given in the following table.

Characteristic (Identification number ) Crossover frequency

Speckled body (3) and scabrous eye (2)

Brown eyes (4) and plexus wings (1)

Plexus wings (1) and scabrous eyes (2)

Speckled body (3) and brown eyes (4)

41 %

4 %

34 %

3 %

-from An Introduction to Genetic Analysis

Jan 98,30


26. Phenotypically, generation III offspring will be

A. female, and each individual will be genotypically different

B. female, and each individual will be genotypically identical

C. 50% male and 50% female, and will be genotypically different D. 50% male and 50% female, and will be gentotypically identical

Jan 98,31

27. If a new disease-causing organism from which lizards have no protection reaches an island where a

uniform population of lizards reproduce only asexually, a likely outcome is that

A. rapid extinction of the lizard population will occur

B. the lizard population will begin to reproduce sexually

C. gene frequencies will change in the lizard population’s gene pool

D. unique individuals in the lizard population will not find a mating partner

Jan 98,2

Numerical Response

10. From the list below, identify the processes that correspond to A, B, C and D in the diagram above 26

Normal Processes

1 Fertilization

2 Meiosis

3 Mitosis


Answer: Process ______ ______ ______ ______

A B C D

Jan 98, 32


28. A logical interpretation that can be drawn from this information is that vitamin A may play a major

role in growth by

A. increasing the amount of mitosis

B. increasing the amount of meiosis

C. decreasing the amount of mitosis

D. decreasing the amount of meiosis

Jan 98,33


29. One homologous pair of chromosomes in a human spermatogonium undergoes nondisjunction during

the first division of meiosis. After meiosis is completed, what number of chromosomes will the four

newly produced cells contain?

A. All four cells will have 23 chromosomes

B. Two cells will have 22 chromosomes, and two cells will have 24 chromosomes

C. One cell will have 22 chromosomes, and three cells will have 24 chromosomes

D. One cell will have 24 chromosomes, and three cells will have 22 chromosomes

Jan 98,34

30. Which expression of chromosome content represents somatic cells in people with trisomy disorders

such as Down syndrome?

A. n – 1

B. n + 1

C. 2n – 1

D. 2n + 1

Many adult newts and salamanders have a remarkable ability to regenerate amputated limbs.

After amputation of a foot, a newt will regenerate the lost foot. However, if a newt has its foot

amputated and receives a particular dosage of vitamin A, the animal grows back a whole new

forelimb, not just the foot!

-from Pietsch

Nondisjunction in meiosis disrupts the chromosome number in the gametes that are produced.

Nondisjunction can occur in either the first or second division of meiosis and results in various

genetic disorders

Jan 98,35

Use the following additional information to answer the next question

31. If a diploid plant and a tetraploid plant, each capable of normal meiosis, were crossed, the chromosome

number in their offspring would be

A. 2n

B. 3n

C. 4n

D. 6n

Jan 98,36


32. This diagram of a cellular structure shows

A. one crossed-over pair of chromosomes

B. two homologous chromosomes

C. one tetrad of chromatids

D. two sister chromatids

Polyploids are organisms with three or more complete sets of chromosomes. If a diploid

organism is 2n, then a triploid is 3n, a tetraploid is 4n, and so on.

All major groups of seed plants have some polyploid members. Plant polyploids are larger

than plants with 2n chromosome number. Plant polyploids with even chromosome numbers

(e.g., 4n) can usually produce pollen and seeds. Plant polyploids with odd chromosome

numbers (e.g., 5n) are nearly always sterile.

Jan 98,37


33. To obtain all the representative DNA of an organism, it would be necessary to collect only

A. an egg

B. a sperm

C. a body cell

D. a cell from each type of body tissue

Jan 98,3

Numerical Response

11. Of the nitrogen-based molecules present in the DNA of sea urchins, 17.5% are cytosine molecules.

Calculate the percentage composition of thymine in sea urchin DNA

(Record your answer as a percentage rounded to one decimal place in the numerical-response

section of the answer sheet.)

Answer: _____________%

Jan 98,38


34. If the traits for seed coat texture and seed coat colour had been located close together on the same

chromosome, Mendel might not have conceptualized

A. gene pairs

B. dominance

C. the Law of Segregation

D. the Law of Independent Assortment

The San Diego Zoo is preserving DNA from hundreds of species by freezing cell samples in

its Centre for the Reproduction of Endangered Species. The cell specimens are often from

individuals in wild populations that are chosen for distinctive characteristics.

-from Vedantam

Gregor Mendel examined the inheritance of two traits in pea plants: seed coat texture and

colour. Seed coat texture can be represented as S-smooth and s-wrinkled, and seed coat

colour can be represented as Y-yellow and y-green. SSYY plants were crossed with ssyy

plants to yield F1 pea seeds that were all smooth and all yellow. By crossing plants grown

from these F1 seeds, Mandel obtained four different phenotypes of F2 seeds:

smooth and green seeds

wrinkled and green seeds

smooth and yellow seeds

wrinkled and yellow seeds

Jan 98,4

Numerical Response

12. The F2 seed phenotype ration that Mendel obtained upon crossing two heterozygous smooth and

yellow F1 individuals would have been ____________.


Answer: ___________ : ____________ : __________ : ___________

smooth and wrinkled and smooth and wrinkled and

green green yellow yellow

Jan 98,39


35. This type of cross is referred to as a

A. test cross

B. monohybrid cross

C. homozygous cross

D. heterozygous cross

Mendel selected two varieties of pea plants from seeds he had grown. One variety of peas

came from a field planted with smooth, yellow seeds. Another variety of peas came from a

field planted with wrinkled, green seeds. These two varieties of peas were crossed to produce

255 plants with smooth and green seeds

268 plants with wrinkled and green seeds

237 plants with smooth and yellow seeds

240 plants with wrinkled and yellow seeds

From the phenotype ratio of the offspring, Mendel deduced that the smooth and yellow parents

had the genotype YySs

Jan 98,40


36. What is the probability of obtaining a black puppy from the following cross?

BbEe x BbEE

A. 9/16

B. 3/16

C. 3/4

D. 1/4

Jan 98,5

Numerical Response

13. Two dogs, each with the genotype BbEe, were crossed. What is the percentage probability that their

offspring would have yellow coat colour?

(Record your answer as a whole number percentage in the numerical-response section of the answer

sheet.)

Answer: __________

Jan 98,41


37. What ratio of wild-type to white-eyed progeny can be expected in each sex if F1 females are crossed to

males of the same genotype as their father?

A. Males – 1:0, females – 1:0

B. Males – 1:1, females – 1:0

C. Males – 0:1, females – 1:1

D. Males – 1:1, females – 1:1

In Labrador retriever dogs, two alleles, B and b, determines whether the coat colour will be

black (B) and brown (b). Black coat colour is dominant. A second pair of alleles, E and e, are

on a separate chromosome from B and b. the homozygous recessive condition, ee, prevents the

expression of either allele B or b and produces a dog with a yellow-coloured coat. Some

examples of genotypes and phenotypes for Labrador retrievers are shown below.

Genotype Phenotype

BBEe black

bbEe brown

Bbee yellow

-from Davol

A recessive allele causes Drosophila to have white eyes instead of wild-type eyes. This eye

colour gene is known to be X-linked. In a cross between homozygous wild-type females

and white-eyed males, all F1 progeny have wild-type eyes.

Jan 98,42


38. The crossover frequency between genes e and r is

A. 3.5 %

B. 2.0 %

C. 1.5 %

D. 0.5 %

Jan 98,43


39. During meiosis, which pair of genes have the best chance of being transferred together to a new cell?

A. Black body and purple eyes

B. Purple eyes and speck body

C. Dumpy wings and purple eyes

D. Dumpy wings and speck body

Crossover Frequencies for Some Genes on Drosophila Chromosome One


White eyes (w) and Facet eyes (f) 1.5 %

White eyes (w) and Echinus eyes (e) 4.0 %

White eyes (w) and Ruby eye (r) 6.0 %

Facet eyes (f) and Echinus eyes (e) 2.5 %

Facet eyes (f) and Ruby eyes (r) 4.5 %

Jan 98,44

40. To determine whether this is an X chromosome or an autosome, a researcher would have to determine

whether these traits are

A. recessive

B. dominant

C. passed from male parents to their male offspring

D. passed from female parents to their male offspring

Jan 98,45


41. Which individual is a known homozygote for blood type?

A. I – 1

B. I – 2

C. II – 2

D. II – 3

Jan 98,46

42. Which of the following rows correctly identifies the genotypes of individuals III – 2 and III – 3?

Row Individual III – 2 Individual III – 3

A. IBi or I

BIB I

Ai

B. IBi I

Ai or I

AI

A

C. IBi or I

BIB I

Ai or I

AI

A

D. IBi I

Ai

Jan 98,6

Numerical Response


14. In this herd, what is the frequency of the heterozygous genotype?

(Record you answer as a value from 0 to 1, rounded to two decimal places, in the numerical-


Answer: __________

Jan 98,7

Numerical Response

15. If two heterozygous sheep mated, what would be the probability of them having a white lamb?



Answer: __________

Jun 98,27


43. Which of the cattle must have a heterozygous genotype for this trait?

A. Cow 2 and calf 2

B. Cow 3 and calf 3

C. The polled bull and cow 1

D. The polled bull and cow 2

Jun 98,28

44. Which of the above cattle could have two possible genotype?

A. Cow 1

B. Cow 2

C. Cow 3

D. The polled bull

In sheep, white wool is a dominant trait and black wool is a recessive trait. In a herd of 500

sheep, 20 sheep have black wool.

In cattle, hornless or polled (P) is dominant over the horned (p) condition. This is an autosomal

trait. The semen of a polled bull is used to artificially inseminate three cows. Cow 1 (horned)

produces a horned calf, cow 2 (polled) produces a horned calf, and cow 3 (polled) produces a

polled calf

Jun 98,29


45. Which of the above parents could have a baby with blood type O?

A. Parents 1 and 3

B. Parents 1 and 4

C. Parents 2 and 3

D. Parents 2 and 4

Jun 98,30

46. The predicted phenotypic ratio for the children of parents 3 is

A. 1/2 type A and 1/2 type B

B. 1/2 type AB and 1/2 type O

C. 1/4 type A, 1/2 type O, and 1/4 type B

D. 1/4 type A, 1/4 type B, 1/4 type AB, and 1/4 type O

Jun 98,31


47. A man has grey-blue eyes and a woman has green eyes. Which eye colour phenotypes would be

possible for children born to this man and woman?

A. Grey-blue and green

B. Dark blue and brown

C. Light brown and dark blue

D. Brown-green flecked and light blue

Four babies were born in a hospital on the same day. Due to a mix-up at the hospital, there was

some confusion as to the identity of the babies.

Mother Father Parents 1 Type A Type O

Parents 2 Type AB Type B

Parents 3 Type AB Type O

Parents 4 Type O Type B

Assume that there are two gene pairs involved in determining eye colour: one codes for

pigment in the front of the iris and the other codes for pigment in the back of the iris.

If the genotype is then the eye colour is

AABB black-brown

AABb dark brown

Aabb brown

AaBB brown-green flecked

AaBb light brown

Aabb grey-blue

AaBB green

aaBb dark blue

aabb light blue

-from Audesirk, 1996

Jun 98,5

Numerical Response

16. If one parent has light brown eyes and the other has dark brown eyes, what is the probability that they

would have an offspring with grey-blue eyes.

(Record your answer as a percentage to three digits in the numerical-response section of the answer

sheet.)

Answer: __________ %

Jun 98,32


48. The inheritance pattern described indicates that this condition is

A. X – linked

B. Y – linked

C. autosomal

D. codominant

Jun 98,33

49. Which of the following pedigrees is not possible for the above condition?

(Note: Carriers have not been identified.)

Scientists have identified a genetic condition that apparently makes some men prone to

impulsive, violent behaviour. A pedigree was drawn highlighting the violent members of a

particular family. It appeared, from the pedigree, that men who displayed this violent

behaviour inherited this condition form their mothers, not their fathers. Further evidence

showed that this was the mode of inheritance.

-from Richardson, 1993

Jun 98,34

Use the additional information to answer the next question

50. The substance affected by the genetic defect could be

A. ACTH

B. acetycholine

C. cholinesterase

D. norepinephrine

Jun 98, 35


51. The difference in expression of pattern baldness in women and men results from the pattern baldness

gene’s being

A. X – linked

B. Y – linked

C. independent of hormonal balance

D. influenced by the level of testosterone

Jun 98,36

52. Parents who do not display pattern baldness have a son who exhibited pattern baldness by the age of

thirty. If they also have a daughter, she has

A. no chance of going bald, as the HBH

B genotype is not possible

B. no chance of going bald, as the HBH

N genotype is not possible

C. a 1 in 3 chance of going bald, as the HBH

N genotype is possible

D. a 1 in 2 chance of going bald, as the HBH

B genotype is possible

Researchers found that the allele that caused violent behaviour was situated on a chromosome in

a region that codes for an enzyme responsible for breaking down neurotransmitters in the brain.

- from Richardson, 1993

Two alleles (HB and H

N) exist for the pattern baldness gene. The H

B allele is dominant in men

but recessive in women. The HN (normal) allele is dominant in females bur recessive in males. A

single HB allele seems to cause pattern baldness only in the presence of the level of testosterone

normally found in adult males. Males will develop pattern baldness with the genotypes HBH

N or

HBH

B; whereas, for females to develop pattern baldness, they must inherit two H

B alleles. It

appears that testosterone destroys or inhibits the production of an enzyme necessary for hair

growth in the hair follicle.

- from Hoffman and Lingna, 1995

Jun 98,6


Numerical Response

17. Use the legend to indicate the order of these genes along a chromosome.


Answer: _____ _____ _____ _____

(There are two possible ways of recording the answer; either will be acceptable.)

Jun 99,32


53. The fact that exposure to sunlight increases melanin production in many humans and produces a tan

demonstrates that

A. some people have mutations that prevent melanin production

B. the expression of some genes is influenced by the environment

C. the environment causes mutations that increase the chance of survival

D. the environment causes mutations that have no effect on the chance of survival

Jun 99,33

54. In the type of albinism described above, because melanin production is controlled by an autosomal

gene, it is expected that

A. males will develop albinism as they mature

B. males will inherit albinism from their mothers

C. albinism will occur more frequently among males than females

D. albinism will occur with equal frequencies among males and females

In Drosophila (fruit flies), the genes for pink eyes, rough eyes, curled wings and hairless bristles are

located on chromosome 3.

Legend

Pink eyes – 1

Rough eyes – 2

Curled wings – 3

Hairless bristles - 4

-from Griffiths, 1993


pink eyes and hairless bristles 21.5

hairless bristles and curled wings 19.5

rough eyes and curled wings 41.1

pink eyes and rough eyes 43.1

rough eyes and hairless bristles 21.6

Melanin pigments range in colour from yellow to reddish –brown to black. The amount and the

colour of melanin in the skin account for differences in human skin coloration.

Albinism is a genetic disorder that results in unpigmented skin and other tissues. About 1 in

20,000 humans has albinism. In humans, it can be caused by an autosomal recessive allele (a). Its

dominant allele (A) results in normal pigmentation.

Jun 99,5


Numerical Response

18. What is the probability that any offspring produced by individuals II-5 and II-6 would have piebald

spotting?

(Record your answer as a value form 0 to 1 rounded to two decimal places in the numerical-

response section on the answer sheet.)

Answer: __________

Jun 99,34


55. A plant heterozygous for both traits was crossed with a plant homozygous recessive for both traits.

What percentage of the offspring produced would be expected to display at least one of the dominant

traits?

A. 25%

B. 50%

C. 75%

D. 100%

In garden peas, the allele for tall plant height (T) is dominant over the allele for short plant

height (t), and the allele for axial flower position (A) is dominant over the allele for terminal

flower position (a). The alleles for plant height and flower position assort independently.

Piebald spotting is a rare human disorder. Although this disorder occurs in all races, piebald spotting is most

obvious in people with dark skin. A dominant allele appears to interfere with the migration of pigment-

producing cells; thus, patches of skin and hair lack pigment, allowing “spots” to form.

Pedigree Chart for Piebald Spotting

Jun 99,6


Numerical Response

19. If an “alligator man” were to marry a woman homozygous for the normal condition, what is the

percentage probability that their children would have ichthyosis?

(Record your answer as a whole number percentage in the numerical-response section on the answer

sheet.)

Answer: __________ %

Jun 99,35


56. The genotypes of the parents to whom this Punnett square applies are

A. heterozygous Band homozygous A

B. heterozygous O and homozygous A

C. homozygous B and heterozygous A

D. heterozygous Band heterozygous A

“Alligator men” or “fish women” were exhibited for their physical abnormalities in fairs or

circuses earlier this century. These people probably suffered from X-linked ichthyosis, which

produces symmetric dark scales on the body. The disease occurs in 1 in 6,000 males and much

more rarely in females. Ichthyosis is likely a recessive disorder.

-from Cummings, 1994

F1 Blood Type Cross

IAI

B I

Ai

IAI

B I

Ai

Jun 99,36


57. Which children could belong to a couple in which the woman has blood type A, N, Rh+ and the man

has blood type O, M, Rh+?

A. Children 1 and 3

B. Children 1 and 4

C. Children 2 and 3

D. Children 2 and 4

Jun 99,7

Use the following to answer the next question

Numerical Response

20. What is the frequency of the recessive -thalassemia allele in the gene pool of this population?

(Record your answer as a value form 0 to 1 rounded to one decimal place in the numerical-response

section on the answer sheet.)

Answer: __________

A, B, M, N, O, and Rh Blood Typing

The alleles for A (I

A) and B (I

B) are codominant, and both are dominant to O (i).

The alleles for M and N are codominant

The allele for Rh+ is dominant to the allele for Rh

-.

Blood groups can be used to determine relationships for a variety for legal and medical

purposes. The following is a list of phenotypes of some children over whom there is a legal

dispute.

Blood Types

Child 1 O MN Rh+

Child 2 A N Rh+

Child 3 A MN Rh-

Child 4 AB MN Rh-

A program to detect carriers of -thalassemia (a mild blood disorder) found the incidence of the

disease to be 4% in a particular population. A recessive allele found on an autosomal chromosome

causes -thalassemia.

Jun 99,37


58. If a man and a woman who are both heterozygous for the alleles HbA and Hb

S have a child, the

probability that the child would not be heterozygous is

A. 0.00

B. 0.25

C. 0.50

D. 0.75

Jun 99,8


Numerical Response

21. The phenotype of III-6 is unknown. What is the probability that this individual is a carrier of the sickle

cell allele?

(Record your answer as a whole number percentage in the numerical-response section on the answer

sheet.)

Answer: ________ %

Sickle cell anemia is caused by the sickle cell allele (HbS) of a gene that contributes to

hemoglobin (Hb) production. The abnormal hemoglobin (hemoglobin-S) produced causes red

blood cells to become deformed and block capillaries. Tissue damage results. Affected

individuals homozygous for the sickle cell gene rarely survive to reproductive age.

Heterozygous individuals produce both normal hemoglobin and a small percentage of

hemoglobin-S. These individuals are more resistant to malaria than individuals who are

homozygous for the allele for normal hemoglobin (HbA). Their red blood cells are prone to

sickling when there is a deficiency of oxygen.

Jun 96,38

Use the following additional information to answer the next two questions

59. What is the frequency of the HbA allele in the human gene pool in western Africa?

A. 0.72

B. 0.85

C. 0.90

D. 0.95

Jun 99,39

60. Which of the following conclusions can be drawn from all the information provided on sickle cell

anemia?

A. The sickle cell gene will eventually disappear because of its interaction with malaria

B. Malaria causes heterozygous individuals to be less fertile than homozygous individuals

C. In Africa, sickle cell anemia will disappear since it is lethal in the homozygous condition

D. In Africa, carriers for sickle cell anemia have an advantage over homozygous individuals

Jun 99,40


61. The technology of transferring a gene from bacterium into a green plant is based on the principle that

A. all genes carry the same genetic information

B. all genes have the same basic chemical components

C. the genotypes of the bacterium and green plant are the same

D. the phenotype of an organism is not altered when one gene is exchanged for another

The malaria-causing microorganism Plasmodium falciparum is injected by mosquitoes into the

bloodstream of humans. Historically, the frequency of the HbS in Africa relates directly to the

presence of malaria-causing organisms. In western Africa, the frequency of the HbS allele in

the gene pool is 0.15. In central Africa, the frequency is 0.10, and in southern Africa the

frequency is 0.05.

A bacterium has been found that produces a form of plastic polyhydroxybutyrate (PHB).

Genes from this bacterium have been transferred into a weed called Arabidopsis thaliana.

These weeds now produce a biodegradable plastic.

- from Poirier, et al., 1997

Jan 00,28


62. These data indicate that, regardless of its branch source, pollen has no effect on the leaf colour of

resulting offspring. A reasonable explanation for this observation is that

A. leaf colour is a codominant trait

B. leaf colour is a dominant – recessive trait

C. cell organelles or cytoplasm are active only in pollen

D. cell organelles or cytoplasm contain genetic information

Jan 00,29


63. With respect to the alleles for flower colour, these results indicate

A. X – linked inheritance

B. gene – linked inheritance

C. dominant – recessive inheritance

D. incomplete dominance inheritance

The flowering plant, Mirabilis jalapa (M.jalapa) may have branches with all white leaves, all

green leaves, and all variegated leaves (leaves with green and white patches) on the same plant.

Leaf colour is dependent on the colour of plastids present in cytoplasm. As in the case of other

plants, pollen (containing sperm nuclei) contributes chromosomes but almost no cytoplasm to the

zygote. The ovule contributes both chromosomes and cytoplasm to the zygote. The following data

of offspring phenotypes were collected from crosses between flowers from various branches.

Source of ovule (female)

Source of pollen (male) White branch Green branch

White branch White offspring Green offspring

Green branch White offspring Green offspring

Variegated branch White offspring Green offspring

-from Griffiths, 1993

Several geneticists studied M. jalapa plants with deep crimson flowers and M. jalapa plants

with yellow flowers. Cross-pollinating these plants produced plants with scarlet-red flowers

(F1 generation).

These F1 plants were allowed to self-pollinate, and the resulting seeds produced M. jalapa

plants with three different flower colours. Data similar to the following were collected for

flower colour:

140 deep crimson

310 scarlet-red

160 yellow

-from Engels, 1975

Jan 00,30


64. The likely genotypes of the P1 plants for these two crosses is represented in row

Row

P1 genotypes

scarlet-red-flowered offspring

P1 genotypes

orange-flowered offspring

A RPR x rr RR x rr

B RPR

P x rr RR x rr

C RPr x R

Pr Rr x Rr

D RPR

P x RR R

PR x Rr

Jan 00,31

65. Which of the following phenotypes is the predicted flower colour of M. jalapa with the genotype RPR?

A. Yellow

B. Orange

C. Crimson

D. Scarlet-red

Jan 00,32


66. Which of the following parental genotypes could produce offspring with the four different colour

patterns?

A. BBYy x BbYy

B. BbYY x Bbyy

C. bbYY x bbyy

D. Bbyy x bbYy

A different variety of homozygous M. jalapa produces flowers that are light crimson. Pure-

breeding genotypes and phenotypes are:

RPR

P – deep crimson

RR – light crimson

rr – yellow

When two pure-breeding P1 plants are cross-pollinated, only scarlet-red-flowered offspring

(RPr) are produced.

When another pair of pure-breeding P1 plants are cross-pollinated, only orange-flowered

offspring (Rr) are produced.

-from Engels, 1975

Feather colour in parakeets is controlled by two genes. For one pigment gene, the B allele

produces blue colour and the b allele does not produce any colour. For the other pigment

gene, the Y allele produces yellow colour and the y allele does not produce any colour. Any

genotype containing at least one B allele and one Y allele will produce a green parakeet.

Jan 00,33

67. Which is the probability of obtaining a blue parakeet when two green heterozygous parakeets are

crossed?

A. 0

B. 3/16

C. 1/4

D. 9/16

Jun 00,33


68. The allele that causes cystic fibrosis most likely results in a faulty amino acid sequence for the

A. channel proteins

B. mucin molecules

C. defensin molecules

D. Pseudomonas bacteria

Cystic fibrosis is a recessive Mendelian trait in the human population. A symptom of cystic

fibrosis is the production of large amounts of mucin protein. New studies indicate that

although the cystic fibrosis condition is present at birth, increased mucin production is

preceded by an infection with the bacterium Pseudomonas aeruginosa. Individuals who are

not affected by cystic fibrosis produce a natural antibiotic, defensin, that kills the

Pseudomonas aeruginosa and eliminates the stimulus for increased mucin production.

Defensin is destroyed by a high chloride content in the tissues of individuals with cystic

fibrosis as a result of faulty chloride-channel proteins.

-from Sternberg, 1997

Jun 00,34

69. In a normal individual, the population of Pseudomonas aeruginosa exhibits which of the following

population growth curves following initial infection of the individual?

Jun 00,35



71. Prior to performing amniocentesis, a genetic counsellor collected pedigree information regarding the

incidence of cystic fibrosis within this family. The row that indicates the genotypes of individuals I-1,

I-2 and II-2 is

Row I-1 I-2 II-2

A. Aa Aa aa

B. AA aa Aa

C. XAY X

AX

A X

aY

D. XAY X

AX

a X

AY

Jun 00,36

70. Down syndrome is a trisomy disorder that can be caused by the presence of three copies of

chromosome 21. Which of the following chromosome combinations identifies Down syndrome?

A. 46 chromosomes consisting of 45 autosomes and 1 sex chromosome

B. 46 chromosomes consisting of 44 autosomes and 2 sex chromosomes

C. 47 chromosomes consisting of 45 autosomes and 2 sex chromosomes

D. 47 chromosomes consisting of 44 autosomes and 3 sex chromosomes

Jun 00,37

71. A genetic abnormality such as Down syndrome can be diagnosed by using the cells obtained during

amniocentesis to create a

A. karyotype

B. therapeutic gene

C. DNA fingerprint

D. recombinant vector

Amniocentesis is a common prenatal procedure used to obtain cell to test for genetic

abnormalities that lead to disorders such as Down syndrome, cystic fibrosis, and hemophilia.

The test is usually offered between the 15th

and 18th

weeks of pregnancy to women who have

an increased risk of having children with genetic abnormalities.

Pedigree of a Family with Cystic Fibrosis

Note: Cystic fibrosis in this family is caused by a recessive allele that is found on

chromosome 7.

Jun 00,38


72. A female hemophiliac marries a man who is not a hemophiliac. The row that indicates the probability

of this couple having a child that is a hemophiliac and the sex that the child would be is

Row Probability Sex of Affected Child

A. 0.25 male

B. 0.25 either female or male

C. 0.50 male

D. 0.50 either male or female

Jun 00,6

Numerical Response

22. A woman who is not a hemophiliac has a father who is a hemophiliac. If this woman marries a man

who is a hemophiliac, what is the probability of them having a hemophiliac son?

Answer: __________

(Record your answer as a value from 0 to 1, round to two decimal places in the numerical-response

section on the answer sheet.)

Jun 00,39


73. The genotypes of the parents of these F1 offspring could be

A. BBDD x bbdd

B. BbDD x bbdd

C. Bbdd x bbDD

D. bbDD x BBdd

Jun 00, 40

74. Plants of the F1 generation are suspected of being heterozygous for both genes. A test cross of

colourless plants with the heterozygote plants should produce a phenotypic ratio in the offspring of

A. 1 : 0

B. 3 : 1

C. 2 : 1 : 1

D. 1 : 1 : 1 : 1

In humans, the allele for normal blood clotting, H, is dominant to the allele for hemophilia, h.

The trait is X-linked.

Two different genes control the expression of kernel colour in Mexican black corn: black

pigment gene B and dotted pigment gene D. Gene B influences the expression of gene D.

The dotted phenotype appears only when gene B is in the homozygous recessive state. A

colourless variation occurs when both genes are homozygous recessive.

After pure-breeding black-pigmented plants were crossed with colourless plants, all of the

offspring were black-pigmented.

-from Griffiths et al., 1993

Jun 00,7

Numerical Response

23. What is the probability of dotted offspring being produced from the test cross described above?

Answer: __________


response section on the answer sheet.)

Jun 00,8

Numerical Response

24. If the total number of offspring produced in the test crosses was 1,024 plants, how many plants would

be expected to be black pigmented?

Answer: _________

(Record your answer as a whole number in the numerical-response section on the answer sheet.)

Jan 01,29


75. The genetic composition of the parents is

A. PpTt and PPTT

B. PPTt and PpTT

C. PpTt and PpTt

D. PpTt and Pptt

Jan 01,30

76. One of the green-stemmed, red-tomato plants was crossed with another tomato plant. One of the

offspring was a purple-stemmed, yellow-tomato plant. If this offspring were crossed with a green-

stemmed, yellow-tomato plant, then the possible phenotype of phenotypes of the offspring would be

A. green-stemmed, yellow-tomato plants

B. green-stemmed, yellow-tomato plants and purple-stemmed, yellow-tomato plants

C. green-stemmed, yellow-tomato plants; purple-stemmed, yellow-tomato plants; and

purple-stemmed, red-tomato plants

D. green-stemmed, yellow-tomato plants; purple-stemmed, yellow-tomato plants; purple-

stemmed, red-tomato plants; and green-stemmed, red-tomato plants

In tomato plants, purple stems (P) are dominant to green stems (p), and red tomatoes (T) are

dominant to yellow tomatoes (t). The two genes are located on separate chromosomes.

A purple-stemmed, red-tomato plant is crossed with a purple-stemmed, yellow-tomato plant.

They produce:

28 purple-stemmed, red-tomato plants

31 purple-stemmed, yellow-tomato plants

11 green-stemmed, red-tomato plants

9 green-stemmed, yellow-tomato plants

Jan 01,31


Gene Loci for a Tomato Plant

From Griffiths et al., 1993

77. During meiosis, which of the following pairs of genes has the greatest chance of being separated by

crossing over?

A. (m) and (d)

B. (ne) and (p)

C. (m) and (lc)

D. (p) and (o)

Jan 01,32


78. The cross-over frequency between genes O and S is

A. 6%

B. 29%

C. 31%

D. 97%

Cross-over frequencies for some genes on a tomato plant:

Genes Cross-Over Frequency

normal leaf (M) and tall plant (D) 12%

normal leaf (M) and normal tomato (O) 33%

normal leaf (M) and simple inflorescence (S) 64%

tall plant (D) and normal tomato (O) 21%

tall plant (D) and simple inflorescence (S) 52%

Jan 01,33


79. In the dihybrid cross between the two black mice, the C allele codes for

A. black colour

B. brown colour

C. colour absent

D. colour present

Jan 01,6

Numerical Response

25. What is the expected phenotypic ratio that results from a cross between two black mice heterozygous

for both genes?

Phenotypic Ratio: __________ : __________ : __________

Coat Colour: Black Brown White

(Record your three-digit answer in the numerical-response section on the answer sheet.)

Jan 01,7

Numerical Response

26. What is the expected phenotypic ratio resulting from a cross between a bbCc female mouse and BbCc

male mouse?

Phenotypic Ratio: __________ : __________ : __________

Coat Colour: Black Brown White

(Record your three-digit answer in the numerical-response section on the answer sheet.)

Jan 01,34


80. If the ram is heterozygous for white wool, the expected phenotypes of the offspring of the farmer’s test

cross would be

A. all black

B. all white

C. 1/2 black and 1/2 white

D. 3/4 black and 1/4 white

Farmers who raise sheep for wool try not to produce offspring with black wool. Black wool is

very brittle and difficult to dye; therefore, white wool is more desirable. If a farmer purchases

a white ram, he will generally carry out a test cross to determine whether the ram is

heterozygous and homozygous for white wool. White wool (W) is dominant to black wool (w).

Jan 02,5


Numerical Response

27. Using the number above, match these descriptions and symbols with the term below to which they

apply

Description or

Symbol Number: __________ ___________ ___________ ___________

Term: gene allele phenotype genotype

(Record all four digits of your answer in the numerical-response section on the answer sheet.)

Jan 02,28

81. What are the genotypes for coat colour of two horses that are predicted to produce offspring in a 1:1

genotypic ratio?

A. Tt and tt

B. Tt and Tt

C. Tobiano and tobiano

D. Tobiano and not tobiano

Description and Symbols Used to Represent One Type of Coat Colour in Horses

1 2 3 4 DNA sequence for coat colour TT, Tt T Tobiano (white spotting pattern)

Tt t Not tobiano (no white spotting

pattern)

Jan 02,6

Numerical Response

28. Given that the diploid number for horses is 64, what is the number of chromosomes found in a horse’s

somatic cell and what is the number of chromosomes found in a horse’s gamete cell?

Number of

Chromosomes: _____ _____ , _____ _____

Cell type: somatic cell gamete cell

(Record all four digits of your answer in the numerical-response section on the answer sheet.)

Jan 02,29

Use the following information to answer the next six questions

82. According to the data above, the relationship among these alleles is such that the

A. black allele is codominant with the chocolate and cinnamon alleles

B. black allele is codominant with the chocolate allele, and the chocolate allele is

codominant with the cinnamon allele

C. black allele is dominant over the chocolate and cinnamon alleles, and the chocolate allele

is dominant over the cinnamon allele

D. black allele is dominant over the chocolate and cinnamon alleles, and the chocolate and

cinnamon alleles are codominant

Cat coat colour results from the interaction of three different genes. A gene for black-based

colours is located on an autosomal chromosome. A gene for red-based colours is located on

the X chromosome. A different gene located on a separate autosomal chromosome

determines pigment density in cat hair.

The black-based gene has three possible alleles: B – black, b – chocolate, and b1 – cinnamon.

If pigmentation in cat hair is dense, the phenotypes listed below are possible

Genotype Phenotype

BB, Bb, Bb1 black

bb, bb1 chocolate

b1b

1 cinnamon

Jan 02,30


83. A blue-coloured female cat is bred with a cinnamon-coloured male cat. The offspring produced are

black-coloured, blue-coloured, chocolate-coloured, and lilac-coloured. The genotypes of the parental

cats are indicated in row

Row Female Cat Male Cat

A. Bb1dd b

1b

1Dd

B. Bb1dd b

1b

1DD

C. Bbdd b1b

1Dd

D Bbdd b1b

1DD

Jan 02,31

84. A black-coloured female cat with the genotype BbDd is bred with a fawn-coloured male cat. The

percentage of their offspring predicted to be chocolate-coloured is

A. 13 %

B. 19 %

C. 25 %

D. 50 %

There are two alleles for the pigment-density gene: dense pigment (D) and dilute pigment (d).

The chart below shows the interaction of two autosomal genes affecting coat colour – the

black-based gene and the density gene.

Black-based pigment gene

Density gene

D_ dd

B_

B_D_

black colour

B_dd

blue colour

bb; bb1

bbD_; bb1D_

chocolate colour

bbdd; bb1dd

lilac colour

b1b1

b1b

1D_

cinnamon colour

b1b

1dd

fawn colour

Jan 02,32

Use the following additional information to answer the next three questions

85. The phenotype of a female cat with genotype XrX

rBb

1 would be

A. a black cat

B. an orange cat

C. an orange-and-black cat

D. an orange, black, and cinnamon cat

Jan 02,33

86. A cinnamon-coloured male cat (XrYb

1b

1) is bred with an orange-coloured female cat (X

RX

RBB). What

possible phenotypes could be produced in the offspring?

A. Tortoiseshell-coloured female cats and orange-coloured male cats

B. Tortoiseshell-coloured female cats, black-coloured female cats, and black-coloured male

cats

C. Cinnamon-coloured male cats, orange-coloured female cats, and tortoiseshell-coloured

female cats

D. Cinnamon-coloured male cats, black-coloured male cats, black-coloured female cats,

orange-coloured female cats, and tortoiseshell-coloured female cats

In cats, red pigmentation is dominant to black pigmentation. The red pigment gene, which

is located on the X chromosome, has two alleles: XR and X

r. Cats with at least on X

R allele

have some orange-coloured hair as a result of having the red-based pigment. Cats with only

Xr alleles have no red-based pigment. Male cats with the X

R allele will be orange. However,

female cats express the genes on only one X chromosome in each cell. This expression is

random. Therefore, an orange-and-black (tortoiseshell) female cat is possible if it is XRX

r.

Some genotypes and their resulting phenotypes are shown below. In all cases, pigment

density is high.

Genotype Phenotype

XRYBb Orange male cat

XrYBb

1 Black male cat

XRX

rBb Orange-and-black female cat (tortoiseshell)

Jan 02,34


87. What is the predicted phenotype of a female cat with genotype XRX

RBb

1dd?

A. Black

B. Orange

C. Cinnamon

D. Cream (light orange)

When the three genes that code for black-based colour, red-based colour, and density combine, they

produce other coat colours in cats.

mendelian geneticscachscomp20.com/bio30/mendel.pdf · 2014-06-19 · mendelian genetics ... if tall...

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